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HW5 - Solutions to Homework 5 Math 293 Section 2.3 Problem...

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Solutions to Homework # 5, Math 293 Section: 2.3 Problem 19 Acceleration due to the motor= 4 ft/s 2 Deceleration due to water resistance = v 2 / 400 ft/s 2 So, the differential equation is: v = 4 - 1 400 v 2 , v (0) = 0 or dv 4 - 1 400 v 2 = dt 1 40 1 - ( v/ 40) 2 = 1 10 dt tanh - 1 ( v 40 ) = t/ 10 + C at t = 0, v = 0. Therefore, C = 0. v ( t ) = 40 tanh( t 10 ) v (10) = 40 tanh(1) = 30 . 46 ft/s v ( ) = 40 tanh( ) = 40 . Problem 28 Given differential equation is dv dt = - GM r 2 = v dv dr = - GM r 2 After integration and using the initial conditions r (0) = r 0 , v (0) = 0, we have v 2 2 = GM 1 r - 1 r 0 1
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dr dt = v = - 2 GM 1 r - 1 r 0 (1) We take the negative square root because v < 0 in descent. Hence, t = - r 0 2 GM r r 0 - r dr using r = r 0 cos 2 θ , t = r 0 2 GM 2 r cos 2 θdθ t = r 3 / 2 0 2 GM (1 + cos 2 θdθ ) t = r 3 / 2 0 2 GM ( θ + 1 2 sin 2 θ ) + C or t = r 3 / 2 0 2 GM ( θ + sin θ cos θ ) + C Using the initial condition t = 0, r = r 0 and θ (0) = 0, we get C = 0. Substituting cos θ = r/r 0 into the above equation, t = r 0 2 GM rr 0 - r 2 + r 0 cos - 1 r r 0 (b) Substitution of G = 6 . 6726 x 10 - 11 , M = 5 . 975 x 10 24 kg, r = R = 6 . 378 x 10 6 m and r 0 = R + 10 6 in the above equation yields t = 510 . 504 sec = 8 . 5 mins (c)
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